Answer
See answers.
Work Step by Step
We find the Compton wavelength shift for a photon scattered from an electron, using Eq. 27–7.
$$\lambda ‘=\lambda + \frac{h}{m_e c}(1-cos \phi)$$
$$\Delta \lambda = \frac{h}{m_e c}(1-cos \phi)$$
Calculate the Compton wavelength of a free electron.
$$\frac{h}{m_e c} = \frac{6.626\times10^{-34}J\cdot s}{(9.11\times10^{-31}kg) (3.00\times10^8 m/s) }=2.424\times10^{-12}m$$
Finally, we obtain this relation.
$$\Delta \lambda = (2.424\times10^{-3}nm)(1-cos \phi)$$
a. Apply the equation for a scattered angle of $45^{\circ}$.
$$\Delta \lambda = (2.424\times10^{-3}nm)(1-cos 45^{\circ})=7.1\times10^{-4}nm$$
b. Apply the equation for a scattered angle of $90^{\circ}$.
$$\Delta \lambda = (2.424\times10^{-3}nm)(1-cos 90^{\circ})=2.4\times10^{-3}nm$$
c. Apply the equation for a scattered angle of $180^{\circ}$.
$$\Delta \lambda = (2.424\times10^{-3}nm)(1-cos 180^{\circ})=4.8\times10^{-3}nm$$